1G.2] 1G. Borelfunctions and isomorphisms 39
If g(x) = α,then
a(n) = k ⇐⇒ d(x,rk)
&(∀s k)[d(x,rs)
Thus if
Bn,k = : α(n) = k},
each g−1[Bn,k] is a Borel subset of X. It follows that for each basic nbhd N = :
α(0) = k0,...,α(n 1) = kn−1} in N,the set
g−1[N]= g−1[B0,k0
] ···
is Borel and g is a Borel function.
Now, easily
α B ⇐⇒ (∀n) d

∀k α(n)
so B is a Π2 0 subset of N. We must refine the construction a bit to get and A with
the same properties, with A a closed set.
Put B in normal form
α B ⇐⇒ (∀n)(∃s)R(α,n,s),
where R is a clopen pointset by 1B.7 and define A N × N by
(α,) A ⇐⇒ (∀n) R
∀k (n)
¬R(α,n,k) .
Clearly A is closed. Moreover, the projection : N × N N, (α,) = α takes A
onto B and is one-to-one on A,since
(α,) A =⇒ (n) = least k such that R(α,n,k).
Hence the composition = takes A onto X and is continuous, one-to-one.
It is trivial to check that the inverse of
f(x) = g(x),n least k such that R
isBorel. Theproofiscompletedbycarrying A to N viasometrivialhomeomorphism
of N with N × N, e.g., the map
The function f of this proof is an example of an interesting class of functions. Let
us temporarily call a function
f : X Y
a good Borel injection if
(1) f is a Borel injection,
(2) there is a Borel surjection
g : Y X
such that g f is the identity on X, i.e.,
= x (x X).
We refer to any such g as a Borel inverse of f.
It will turn out that every Borel injection is a good Borel injection. This is a special
case of a fairly difficult theorem which we will prove in 2E and again in Chapter 4.
Here we only need show that enough good Borel injections exist.
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